A bullet is shot into a block (and embeds itself into the block) that is on a ro

Question

A bullet is shot into a block (and embeds itself into the block) that is on a rough tabletop. The bullet has mass .0045 kg, the bullet has initial speed of 200 m/s, the block has mass 1.5 kg, and the block/bullet comes to rest after sliding .23 m on the table.

12) What is the initial momentum of the bullet?

0.9 kg-m/s

16 kg-m/s

79 kg-m/s

178 kg-m/s

220 kg-m/s

13) What is the speed of the block/bullet immediately after the bullet embeds itself in the block?

.2 m/s

.3 m/s

.4 m/s

.5 m/s

.6 m/s

14) What is the magnitude of the work done by the frictional force on the block as it slides?

.11 J

.19 J

.27 J

.32 J

.38 J

15) What is the coefficient of friction between the block and the table?

.02

.08

.14

.25

.36

Explanation / Answer

Q12. initial momentum of the bullet=mass*speed

=0.0045*200=0.9 kg.m/s

so first option is correct.

Q13. conserving momentum along the direction of original motion,

speed of the block bullet system=initial momentum of the bullet/mass of block bullet system

=0.9/(1.5+0.0045)

=0.59821 m/s

=0.6 m/s (approx)

hence fifth option is correct.

Q14. work done=loss of kinetice enrgy after the bullet is embedded

=0.5*mass of bullet block system*speed of bullet block system^2

=0.5*(1.5+0.0045)*0.59821^2

=0.2692 J
=0.27 J

hence third option is correct.

Q15.magnitude of work done=friction force*distance covered

==>0.2692=friction force*0.23

==>friction force=1.1704 N

==>friction coefficient * normal force=1.1704

==>friction coefficient*(1.5+0.0045)*9.8=1.1704

==>friction coefficient=1.1704/((1.5+0.0045)*9.8)

=0.079381

=0.08 (approx)

second option is correct.

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